# An interesting property of curves $V:$ $x^3$ + $y^3$ = A$z^3$

Let $V$ be the elliptic curve V: $x^3$+ $y^3$ = A$z^3$ where A > 2 is cube free natural number. A conjugate quadratic point of $V$ is one of the form $(a + b\sqrt d, a - b\sqrt d, c)$ (note that all quadratic point $(x, y, z)$ of $V$ can be reduced to have $z$ rational integer multiplying by the conjugate of $z$). I have found the result which follows and I am interested in knowing the opinion and remarks or possible objections of readers and mainly I expect for another proof.

Prove that $V(\mathbb Q)$ has no rational points distinct of $(1, -1, 0)$ if and only if all the quadratic points of $V$ are conjugates.

EXAMPLE.- This is true even with $x^3$ + $y^3$ = 2$z^3$ in which the only rational points are $(1, 1, 1)$ and $(1, -1, 0)$. One has here, with $A = 2$, the point $(4 +2\sqrt{-11}, -1 + \sqrt{-11}, -6)$ which is not conjugate because of the rational point $(1, 1, 1)$.

• Care to share your existing proof? Commented May 4, 2015 at 16:01
• @Justaskin Not at all. I want to say thanks to the reader who has edited the "good" Q (he have erased his name). Commented May 4, 2015 at 16:09
• I solved a similar equation. Used this approach. math.stackexchange.com/questions/1114766/… Can anyone suggest better? Commented May 4, 2015 at 16:19
• @individ You can go from your equation to a V here with A = ab but the solutions (X, Y, Z) here will be polynomials of the 9 degree in your original solutions (x, y, z). Besides here is a question about quadratic points. Commented May 4, 2015 at 16:34
• @individ The A for come from your equation to a V here is a$b^2$ and not the ab above. All the other is true. Anyway, come to the curves V is not a good affair to solve your equation. Commented May 4, 2015 at 17:01